Solve the following system equations by the substitute method 10x+10y=1 , x=y-3

Since one equation already has one variable isolated(x), it gives you a "value" to input into the other equation to use to solve for the other variable(y).
10(y-3) + 10y = 1
10y - 30 + 10y = 1
20y - 30 = 1
20y = 31
y = 31/20 or 1.55
Now you can use this value to solve for x in one of the original equations.
x = (31/20) - 3
x = -29/30 or -1.45
Hope this helps!